A solenoid has a cross-sectional area of 2.60 10-4 m2, consists of 600 turns per meter, and carries a current of 0.7 A. A 10 turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 0.8 resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.11 s. Find the average current induced in the coil.
To find the average current induced in the coil, we can use Faraday's Law of electromagnetic induction.
Faraday's Law states that the induced electromotive force (EMF) is equal to the negative rate of change of magnetic flux through a loop of wire.
The formula for EMF is given by:
EMF = -N*(dΦ/dt)
Where:
EMF is the induced electromotive force,
N is the number of turns in the coil,
dΦ/dt is the rate of change of magnetic flux through the coil.
First, let's find the rate of change of magnetic flux through the coil. The magnetic flux (Φ) is given by:
Φ = B * A
Where:
B is the magnetic field strength,
A is the cross-sectional area of the coil.
To find B, we can use Ampere's Law, which states that the magnetic field strength inside a solenoid is directly proportional to the current and the number of turns per unit length:
B = μ₀ * (N₀ * I) / L
Where:
μ₀ is the permeability of free space (4π * 10^-7 T·m/A),
N₀ is the number of turns per unit length of the solenoid,
I is the current in the solenoid,
L is the length of the solenoid.
Given:
N₀ = 600 turns/m,
I = 0.7 A.
We need to determine L, the length of the solenoid. Unfortunately, this information is not provided in the given data. We can proceed by assuming a reasonable value for L or considering an average value for solenoids.
Assuming L = 1 m (a reasonable value for a solenoid), we can plug the given values into the equation for B:
B = (4π * 10^-7 T·m/A) * (600 turns/m * 0.7 A) / 1 m
B ≈ 1.32 × 10^-3 T
Now, let's find the rate of change of magnetic flux through the coil:
dΦ/dt = d(B * A)/dt = A * dB/dt
Given:
A = 2.60 × 10^-4 m²
The change in magnetic field strength over time (dB/dt) can be calculated using the given data. The current in the solenoid dies to zero, so the rate of change of current (di/dt) can be determined as:
di/dt = -I / t
Where:
I is the initial current in the solenoid,
t is the time for the current to die to zero.
Given:
I = 0.7 A,
t = 0.11 s.
Substituting the values:
di/dt = -0.7 A / 0.11 s
di/dt ≈ -6.36 A/s
Now, let's find dB/dt using the relationship between dB/dt and di/dt:
dB/dt = (μ₀ * N₀ * di/dt) / L
Given:
N₀ = 600 turns/m,
L = 1 m.
Substituting the values:
dB/dt = (4π * 10^-7 T·m/A) * (600 turns/m * -6.36 A/s) / 1 m
dB/dt ≈ -1.52 × 10^-4 T/s
Finally, let's calculate the rate of change of magnetic flux through the coil:
dΦ/dt = A * dB/dt
dΦ/dt = (2.60 × 10^-4 m²) * (-1.52 × 10^-4 T/s)
dΦ/dt ≈ -3.95 × 10^-8 Vs
Now, let's find the average current induced in the coil:
EMF = -N * (dΦ/dt)
Given:
N = 10 turns
Substituting the values:
EMF = -10 turns * (-3.95 × 10^-8 Vs)
EMF ≈ 3.95 × 10^-7 Vs
The EMF induced in the coil is equal to the average current induced multiplied by the resistance:
EMF = I_avg * R
Given:
R = 0.8 Ω
Solving for the average current induced in the coil:
I_avg = EMF / R
I_avg = (3.95 × 10^-7 Vs) / 0.8 Ω
I_avg ≈ 4.94 × 10^-7 A
Therefore, the average current induced in the coil is approximately 4.94 × 10^-7 A.